LeetCode in Kotlin

3239. Minimum Number of Flips to Make Binary Grid Palindromic I

Medium

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make either all rows palindromic or all columns palindromic.

Example 1:

Input: grid = [[1,0,0],[0,0,0],[0,0,1]]

Output: 2

Explanation:

Flipping the highlighted cells makes all the rows palindromic.

Example 2:

Input: grid = [[0,1],[0,1],[0,0]]

Output: 1

Explanation:

Flipping the highlighted cell makes all the columns palindromic.

Example 3:

Input: grid = [[1],[0]]

Output: 0

Explanation:

All rows are already palindromic.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m * n <= 2 * 105
• 0 <= grid[i][j] <= 1

Solution

import kotlin.math.min

class Solution {
    fun minFlips(grid: Array<IntArray>): Int {
        val m = grid.size
        val n = grid[0].size
        var rowFlips = 0
        for (i in 0 until m / 2) {
            for (j in 0 until n) {
                val sum = grid[i][j] + grid[m - 1 - i][j]
                rowFlips += min(sum, (2 - sum))
            }
        }
        var columnFlips = 0
        for (j in 0 until n / 2) {
            for (ints in grid) {
                val sum = ints[j] + ints[n - 1 - j]
                columnFlips += min(sum, (2 - sum))
            }
        }
        return min(rowFlips, columnFlips)
    }
}

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